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GrogVix [38]
3 years ago
9

A spherical exercise ball has a maximum diameter of 30 inches when filled with air. The ball was completely empty at the start,

and an electric air pump is filling it with air at the rate of 1600 cubic Inches per minute. The formula for the volume of a sphere is 4pi (r^3/3)
Part A Enter an equation for the amount of air still needed to fill the ball to its maximum volume, y, with respect to the number of minutes the pump has been pumping air into the ball, X.

Part B Enter the total amount of air, in cubic inches, still needed to fill the ball after the pump has been running for 4 minutes.

Part C Enter the estimated number of minutes it takes to pump up the ball to its maximum volume.
Mathematics
1 answer:
Alex73 [517]3 years ago
8 0

Answer:

A) A(t)  =  4500*π  -  1600*t

B) A(4)  =  7730 in³

C) t  =  8,8 sec

Step-by-step explanation:

The volume of the sphere is:

d max  =  30   r max  =  15 in

V(s) =  (4/3)*π*r³      V(s)  =   (4/3)*π* (15)³

V(s) = 4500*π

A)  Amount of air needed to fill the ball A(t)

A(t)  = Total max. volume of the sphere - rate of flux of air * time

A(t)  =  4500*π  -  1600*t    in³

B) After 4 minutes

A(4)  =  4500*π  - 6400

A(4)  =  14130 - 6400

A(4)  =  7730 in³

C)  A(t)  =  4500*π  -  1600*t  

when  A(t) = 0   the ball got its maximum volume then:

4500*π -  1600*t  =  0

t  =   14130/1600

t  =  8,8 sec

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A rectangular box with a volume of 272ft^3 is to be constructed with a square base and top. The cost per square foot for the bot
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Answer:

The dimensions of the box is 3 ft by 3 ft by 30.22 ft.

The length of one side of the base of the given box  is 3 ft.

The height of the box is 30.22 ft.

Step-by-step explanation:

Given that, a rectangular box with volume of 272 cubic ft.

Assume height of the box be h and the length of one side of the square base of the box is x.

Area of the base is = (x\times x)

                               =x^2

The volume of the box  is = area of the base × height

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Therefore,

x^2h=272

\Rightarrow h=\frac{272}{x^2}

The cost per square foot for bottom is 20 cent.

The cost to construct of the bottom of the box is

=area of the bottom ×20

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The cost per square foot for top is 10 cent.

The cost to construct of the top of the box is

=area of the top ×10

=10x^2 cents

The cost per square foot for side is 1.5 cent.

The cost to construct of the sides of the box is

=area of the side ×1.5

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Putting the value of h

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Differentiating with respect to x

C'=60x-\frac{1632}{x^2}

Again differentiating with respect to x

C''=60+\frac{3264}{x^3}

Now set C'=0

60x-\frac{1632}{x^2}=0

\Rightarrow 60x=\frac{1632}{x^2}

\Rightarrow x^3=\frac{1632}{60}

\Rightarrow x\approx 3

Now C''|_{x=3}=60+\frac{3264}{3^3}>0

Since at x=3 , C''>0. So at x=3, C has a minimum value.

The length of one side of the base of the box is 3 ft.

The height of the box is =\frac{272}{3^2}

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The dimensions of the box is 3 ft by 3 ft by 30.22 ft.

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Answer:

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Answer:

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